Real Analysis query!

Hi,
I'm for once attempting to do the practise questions for the subject but confused already just looking at it!
Anyone able to help with the following?

Show the following:

oo
x~ (2)[SUM ((-1)^(n-1))((sin nx)/n), -PI<x<PI
n=1

oo
x^2~ ((PI^2)/3)+ (4)[SUM ((-1)^n)((cos nx)/(n^2))]
n=1

I'm thinking the problems need the same method so if I could get a start on the first I might be able to figure out the second? It won't leave my oo and n=1 sign where I want them so they are supposed to go above and below the SUM if anyone is wondering

Hi,
I'm for once attempting to do the practise questions for the subject but confused already just looking at it!
Anyone able to help with the following?
oo
x=(approx) SUM 2(-1)^n-1(sin nx)/n, -PI<x<PI
n=1

oo
x^2= (approx) SUM PI^2/3+ 4(-1)^n(cos nx)/n^2
n=1

I'm thinking the problems need the same method so if I could get a start on the first I might be able to figure out the second?

1. What is the question?

2. Add sufficient brackets to make the meaning of the expressions clear.

Presumably these are exercises in calculating Fourier series? For the first one, the function f(x)=x is an odd function, so all the Fourier cosine coefficients will be zero. You calculate the sine coefficients like this:

Okay, the general setting for this is that the Fourier series of a function f(x) on the interval [–π,π] is . The cos coefficients are given by the formula , and the sin coefficients are given by the formula . If the function is even then all the sin coefficients are zero. If it is odd then all the cos coefficients are zero.

The function f(x)=x^2 is even, so all the sin coefficients are zero. For the cos coefficients, you need to evaluate the integral in the formula (you'll need to integrate by parts twice).

Originally Posted by musicmental85

Still stuck on the second if anyone has any ideas. I don't think you can simple square it or maybe you can?

I'm not quite sure what's meant here. If you're wondering whether you can get the Fourier series for x^2 by somehow squaring the series for x, the amswer is definitely NO! You need to work out the integrals to find the coefficients, as above.